Problem 90 - Calculus Single Variable

Question

Suppose that \(f\) is continuous on \([a, b] .\) let \(\delta=(b-a) / 2\) and \(\mu=(a+b) / 2\). Use a substitution to show that $$ \int_{a}^{b} f(u) d u=\delta \int_{-1}^{1} f(\delta x+\mu) d x $$

Step-by-Step Solution

Answer

Use the substitution \( u = \delta x + \mu \) to transform the integral bounds and differential, resulting in the desired integral form.

Step 1: Understanding the Problem

We need to show that the integral of a function \( f(u) \) from \( a \) to \( b \) can be transformed into an integral from \( -1 \) to \( 1 \) using a substitution method.

Step 2: Define Required Substitution Variables

Let us define \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \), which define the transformation we need for changing the interval from \( [a, b] \) to \( [-1, 1] \).

Step 3: Perform the Substitution

First, substitute \( u = \delta x + \mu \). Then differentiate with respect to \( x \): \( du = \delta dx \).

Step 4: Change Integration Limits

Since we have \( u = \delta x + \mu \), when \( u = a \), \( x = -1 \) (because \( a = \mu - \delta \)). When \( u = b \), \( x = 1 \) (since \( b = \mu + \delta \)).

Step 5: Substitute in Original Integral

Substitute \( u = \delta x + \mu \) and \( du = \delta dx \) into the original integral: \[ \int_{a}^{b} f(u) \, du = \int_{-1}^{1} f(\delta x + \mu) \delta \, dx \]

Step 6: Factor and Simplify

Factor out the constant \( \delta \) from the integral:\[ \int_{a}^{b} f(u) \, du = \delta \int_{-1}^{1} f(\delta x + \mu) \, dx \] This shows the desired transformation.

Key Concepts

Substitution Method

The substitution method is a powerful technique for simplifying integrals. It works by replacing a variable in the integral with another variable, usually to transform the integration into a simpler or more recognizable form. In our specific problem, we're substituting the variable \( u \) with \( \delta x + \mu \).

This substitution simplifies the process by aligning the limits of integration with a more symmetric interval \([-1, 1]\). It essentially "re-labels" the input values, which can make solving the integral more convenient.

Here's how it applies in our example:

• We started with the integral \( \int_{a}^{b} f(u) \, du \).
• By substituting \( u = \delta x + \mu \) (where \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \)), we adjust the formula accordingly.
• This leads to the new integral representation \( \int_{-1}^{1} f(\delta x + \mu) \, \delta \, dx \), demonstrating the utility and elegance of the substitution method.

Change of Interval

The change of interval is a specific application of the substitution method. It modifies the range of integration to transform an integral into a more manageable form. In the given problem, the original interval \([a, b]\) is changed to \([-1, 1]\).

This is achieved by recognizing that any shifted interval \([a, b]\) can be transformed using \( \delta = \frac{b-a}{2} \) and \( \mu = \frac{a+b}{2} \). Here's how it works:

• As the substitution \( u = \delta x + \mu \) is applied, the limits of integration move.
• With \( u = a \) leading to \( x = -1 \) and \( u = b \) giving \( x = 1 \), the integral interval is now symmetric around the origin.
• This symmetric interval helps in simplifying certain types of functions, especially those like polynomials or sine and cosine functions.

Using this clever change, computations often become easier, and sometimes more geometric insights into the problem are revealed.

Continuous Function

A continuous function is one that has no breaks, jumps, or gaps in its domain. For integration, especially definite integrals, continuity is crucial because it ensures the integral is well-defined over the interval. In our problem, the function \( f(u) \) is continuous on \([a, b]\).

Continuity allows us to perform operations like substitution and change of intervals without worrying about discontinuities that could disrupt these transformations.

In practical terms:

• A continuous function behaves predictably, allowing us to apply limits and transformations smoothly.
• If \( f \) were not continuous, the integral might encounter undefined sections or infinities that would make calculation difficult or impossible.

Thus, one fundamental takeaway is ensuring that \( f(u) \) maintains its continuity across the original or transformed interval, which justifies the changes we make using the substitution method.

Definite Integrals

A definite integral calculates the "net area" under a curve within specified bounds on the x-axis. For example, \( \int_{a}^{b} f(u) \, du \) evaluates the area from \( a \) to \( b \).

Definite integrals are a central concept in calculus and focus on finding total accumulation or change over an interval.

In our substitution example:

• The transformation from \( \int_{a}^{b} f(u) \, du \) to \( \delta \int_{-1}^{1} f(\delta x + \mu) \, dx \) is a re-expression but still computes this net area.
• Despite different integration limits, both expressions represent the same quantity — emphasizing that the value remains unchanged by substitution, only the integration process does.

This powerful property of definite integrals is part of what makes integral calculus so useful in a variety of scientific and engineering applications.